Additive Subset Picking

How can we choose a subset of $\{1,\ldots,n\}$ with as many elements as possible such that the sum of any two elements is distinct?

Formally, we want a subset $A\subseteq S=\{1,\ldots,n\}$ such that for all distinct unordered pairs
$\{a,b\} \ne \{c,d\},\, a+b\ne c+d$.
If we have two pairs of numbers such that $a-b=x$ and $c-d=x$ where $x\ne 0$, then we would have $a+d=b+c$, and thus
the difference between any two numbers in our subset is unique.

Of note, our valid subsets will be the elements $x \le n$ of
$B_2$ sequences, and are thus
Sidon sets. I'll also use the term
sumset for the set of pairwise sums.
Our problem is the exact one of finding optimal Golomb rulers.

A Sidon set chosen from $\{1,\dots,n\}$ can be shifted to $\{1+d,\ldots,n+d\}$ by an integer $d$, and the resulting
set is still a Sidon set with its sumset shifted by $2d$. Additionally, the set of negated elements of a Sidon set is
still a Sidon set, but with its sumset negated. Thus if $A$ is a Sidon set chosen from $\{1,\dots,n\}$,
its negated pair $B=\{n-x+1 \mid x\in A\}$ is also a Sidon set in $\{1,\dots,n\}$. Unless otherwise specified, I'll
consider Sidon sets here to have their least element be $1$, and I'll additionall call them canonical if its in-order
list of elements is lexicographically before those of its negated pair. This may change in the future,
given any additional symmetries or contraints. For example, $\{4,6,7\}$ is equivalent to
$\{1,3,4\}$, and its canonical form is $\{1,2,4\}$.

The naive approach starts with $\{1\}$ and then counts up and adding a number only if it doesn't introduce
a duplicate pairwise sum. This results in $\{1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, \ldots\}$, known as the
Mian-Chowla sequence
(OEIS A005282). Sidon sets from this sequence are not necessarily the largest;
$\{1,2,5,7\}$ is larger than the Mian-Chowla subsequence $\{1,2,4\}$ for Sidon subsets of $\{1,\ldots,7\}$.